A particle oscillates along the $$ x $$-axis according to the law, $$ x(t)=x_0 \sin ^{2}\left(\frac{t}{2}\right) $$ where $$ x_0 = 1 m $$. The kinetic energy $$ (K) $$of the particle as a function of $$ x $$ is correctly represented by the graph

Given

$$x(t)=x_0\sin^2\left(\frac{t}{2}\right)$$

Use identity

$$\sin^2\theta=\frac{1-\cos2\theta}{2}$$

So

$$x=\frac{x_0}{2}(1-\cos t)$$

Rearrange:

This is SHM about mean position

$$\frac{x_0}{2}$$

with amplitude

$$A=\frac{x_0}{2}$$

Since

$$x_0=1$$

amplitude is

Now in SHM,

$$K=\frac{1}{2}k(A^2-y^2)$$

where displacement from mean is

$$y=x-\frac{1}{2}$$

So

$$K\propto A^2-\left(x-\frac{1}{2}\right)^2$$

Substitute

$$A=\frac{1}{2}$$

$$K\propto\frac{1}{4}-\left(x-\frac{1}{2}\right)^2$$

Expanding,

$$K\propto x-x^2$$

This is a downward-opening parabola.

It is zero at

x=0

and

x=1

and maximum at

So correct graph is

• inverted parabola
• touches x-axis at 0 and 1
• maximum at $$x=1/2$$